AP Physics Unit

Unit 1 - Vectors and Kinematics

Advanced

Mathematical

GQ

Suppose the water at the top of Niagara Falls has a horizontal speed of 2.7 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 75° angle below the horizontal?

5.18 m

0

The water initially has a horizontal velocity of 2.7 m/s. As it falls, gravity accelerates it downward, changing the direction of its velocity vector. We need to find the point where the angle of this vector with the horizontal is 75°.

To solve this, we use the relationship between the horizontal velocity (v_x), the vertical velocity (v_y), and the angle (\theta).

Step Formula Derivation Reasoning
1.1 \tan(\theta) = \frac{v_y}{v_x} The tangent of the angle is the ratio of the vertical velocity to the horizontal velocity.
1.2 \tan(75^\circ) = \frac{v_y}{2.7 , \text{m/s}} Substitute \theta = 75^\circ and v_x = 2.7 , \text{m/s}.
1.3 v_y = 2.7 , \text{m/s} \times \tan(75^\circ) Solve for v_y.
2.1 v_y = gt Vertical velocity at time t under gravity.
2.2 t = \frac{v_y}{g} Solve for time t.
3.1 y = \frac{1}{2} g t^2 Vertical distance fallen in time t.
3.2 y = \frac{1}{2} g \left( \frac{v_y}{g} \right)^2 Substitute t from step 2.2.
3.3 y = \frac{1}{2} g \left( \frac{2.7 , \text{m/s} \times \tan(75^\circ)}{g} \right)^2 Substitute v_y from step 1.3.

We will now calculate these steps.

The vertical distance below the edge of Niagara Falls, where the velocity vector of the water points downward at a 75° angle below the horizontal, is approximately 5.18 , \text{m}.

Need Help On This? Ask Phy To Explain.

Phy can also check your working. Just snap a picture!

Simple Chat Box
NEW Smart Actions

Topics in this question

Discover how students preformed on this question | Coming Soon

Discussion Threads

Leave a Reply

5.18 m

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Sign In to View Your Questions

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!
Made By Nerd-Notes.com
KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: \text{5 km}

  2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

  3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

  4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

Error Report

Sign in before submitting feedback.

Phy Pro

The most advanced version of Phy. Currently 50% off, for early supporters.

$11.99

per month

Billed Monthly. Cancel Anytime.

Trial  –>  Phy Pro

You can close this ad in 15 seconds.

Ads display every few minutes. Upgrade to Phy Pro to remove ads.

Jason here! Feeling uneasy about your next physics test? We will help boost your grade in just two hours.

We use site cookies to improve your experience. By continuing to browse on this website, you accept the use of cookies as outlined in our privacy policy.